Q&A - PSLE Math

[MathIzzzFun]

P5G:

Please help.


Paya Lebar Methodist Girls' Prelim Question 15.
http://www.orlesson.org/orp/09Ma/2009-P6-Math-SA2-PLMGS.pdf


The following pattern is created with 4 letters, A, B, C and D.

ABCDAABBCCDDAAABBBCCCDDD.................?
..............................................24th 230th

Following the above pattern, what is the 230th letter?

(1) A
(2) B
(3) C
(4) D


Given answer is (1).


TIA.

Hi

The pattern can be grouped as follows:
group 1 (each letter appearing once) = ABCD
group 2 (each letter appearing twice) = AABBCCDD
and so forth

The total number of letters in each group will then be 4x1, 4x2, 4x3 etc..

To find the 230th letter, divide 230 by 4 = 57.5, so we need to find the group number n such that 1+2+3+4+...n is less than or equal to 57.5

1+2+3+4+5...+10 = 55

So, we know that 230th letter will be in group 11 ie each letter appearing 11 times,

4 x 55 = 220 (total number of letters in group 1 to group 10), 230 - 220 = 10, since A is the first letter, and each letter appears 11 times, the 10th letter in group 11 or the 230th letter is an \"A\"

cheers.

[fxchow]

Hi,


Can someone help in this question from Raffles Girls'Prm Sch-Prelim-Q18. TIA

There are more pupils in Sch A than Sch B.
30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B.

a) How many pupils are there in Sch B?
b) How many percent less pupils are there in Sch B than Sch A?
Leave your answer as fraction in the simplest form.

[small]

Hi fxchow,


HTH but I am not so sure on my answer for part (b).

a)
Given that 30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
0.3A - 45 = 0.4B --> x10

3A = 4B + 450 ----- equation (1)


If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B
0.9A – 200 = 0.1A + 1B
0.8A = 1B + 200 --> x10

8A = 10B + 2000 ----- equation (2)


We will need to eliminate A to get B;

equation (1) x 8:
24A = 32B + 3600

equation (2) x 3:
24A = 30B + 6000

32B + 3600 = 30B + 6000
2B – 6000 – 3600
2B = 2400
1B = 1200

There are 1200 pupils in Sch B


b)
from equation (1) :

3A = 4 X 1200 + 450
3A = 4800 + 450
3A = 5250
1A = 1750 (100%)

The different between School A and School B is:
1750 – 1200 = 550

1750 = 100%
550 = 550/1750 x 100%
= 1100 / 35 %
= 31 3/7%


There are 31 3/7% percent less pupils in Sch B than Sch A

[MathIzzzFun]

fxchow:

Hi,


Can someone help in this question from Raffles Girls'Prm Sch-Prelim-Q18. TIA

There are more pupils in Sch A than Sch B.
30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B.

a) How many pupils are there in Sch B?
b) How many percent less pupils are there in Sch B than Sch A?
Leave your answer as fraction in the simplest form.

hi fxchow,

hope this helps

http://www.flickr.com/photos/62167097@N02/5842021695/in/photostream

cheers.

[fxchow]

small:

Hi fxchow,


HTH but I am not so sure on my answer for part (b).

a)
Given that 30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
0.3A - 45 = 0.4B --> x10

3A = 4B + 450 ----- equation (1)


If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B
0.9A – 200 = 0.1A + 1B
0.8A = 1B + 200 --> x10

8A = 10B + 2000 ----- equation (2)


We will need to eliminate A to get B;

equation (1) x 8:
24A = 32B + 3600

equation (2) x 3:
24A = 30B + 6000

32B + 3600 = 30B + 6000
2B – 6000 – 3600
2B = 2400
1B = 1200

There are 1200 pupils in Sch B


b)
from equation (1) :

3A = 4 X 1200 + 450
3A = 4800 + 450
3A = 5250
1A = 1750 (100%)

The different between School A and School B is:
1750 – 1200 = 550

1750 = 100%
550 = 550/1750 x 100%
= 1100 / 35 %
= 31 3/7%


There are 31 3/7% percent less pupils in Sch B than Sch A

Hi Small,

Thank you for your solution, your answer b is correct.

[fxchow]

MathIzzzFun:

fxchow:

Hi,


Can someone help in this question from Raffles Girls'Prm Sch-Prelim-Q18. TIA

There are more pupils in Sch A than Sch B.
30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B.

a) How many pupils are there in Sch B?
b) How many percent less pupils are there in Sch B than Sch A?
Leave your answer as fraction in the simplest form.

hi fxchow,

hope this helps

http://www.flickr.com/photos/62167097@N02/5842021695/in/photostream

cheers.

Hi MathIzzzFun,

Thank you for your model drawing, it's great.

fxchow

[MathIzzzFun]

fxchow:

MathIzzzFun:

[quote=\"fxchow\"]Hi,


Can someone help in this question from Raffles Girls'Prm Sch-Prelim-Q18. TIA

There are more pupils in Sch A than Sch B.
30% of the pupils in Sch A is 45 more than 40% of the pupils in Sch B.
If 10% of the pupils in Sch A leaves to join Sch B, there will be 200 more pupils in Sch A than Sch B.

a) How many pupils are there in Sch B?
b) How many percent less pupils are there in Sch B than Sch A?
Leave your answer as fraction in the simplest form.

hi fxchow,

hope this helps

http://www.flickr.com/photos/62167097@N02/5842021695/in/photostream

cheers.

Hi MathIzzzFun,

Thank you for your model drawing, it's great.

fxchow[/quote]u r welcome

cheers.

[ozora]

ksi:

http://i54.tinypic.com/xprbkl.jpg\">

thanks KSi

[corneyAmber]

you are welcome ozora.

[stary]

Hi all,


Have 2 questions that I need help with…

1.
A and B start cycling at the same time from the same place.
If they cycle in the opposite direction, they will be 37 2/5 km apart after 55 min.
If they cycle in the same direction, A will be 2 1/5 km ahead of B after 55 min.
(a) Find the diff. in their average speed.
(b) Find the average speed of the slower cyclist.(Express in decimals)

2.
At 2pm, Melvin started from Vivo City and travelled towards Sun City at an average speed of 78km/h. At 3:12pm, he went past Andrew who was travelling from Sun City to Vivo City. At 3:42pm, the distance between Melvin and Andrew was 75 km.
(a) What was Andrew’s speed?
(b) At what time did Andrew reach Vivi City?


Your help is GREATLY appreciated by me.

Best regards,
Stary